1. We need the number of onto (surjective) functions $f:R\to S$ where $R=\{a,b,c,d,e\},\qquad S=\{1,2,3,4\}$ and additionally $f(a)\ne 1.$

2. Since $|R|=5$ and $|S|=4$, for a function to be onto, exactly one value in $S$ must have two preimages and the other three values must have one preimage each.

So the pattern of sizes of preimage sets is: $2,1,1,1.$

3. Count onto functions with the restriction $f(a)\ne 1$. We split into cases depending on the value of $f(a)$.

Since $f(a)\ne 1$, we must have $f(a)\in \{2,3,4\}.$ So there are $3$ symmetric choices. We count for one fixed value and multiply by $3$.

Let us fix $f(a)=2.$ Now count onto functions under this condition.

4. Since the function is onto, values $1,3,4$ must each appear at least once among $b,c,d,e$. Also value $2$ is already taken by $a$, so among $b,c,d,e$, either:

• none maps to $2$, giving preimage sizes $1,1,1,2$, or
• exactly one maps to $2$, giving preimage sizes $2,1,1,1$.

Thus among $b,c,d,e$, we must assign values so that $1,3,4$ each appear, and $2$ may appear at most once.

5. Count assignments of $b,c,d,e$.

We have 4 elements $b,c,d,e$ to assign. To ensure onto, $1,3,4$ must all appear at least once. Since there are 4 elements, exactly one of the values among $\{1,2,3,4\}$ is repeated among the full 5-element domain.

With $f(a)=2$, two possibilities:

Case 1: No one among $b,c,d,e$ maps to $2$

Then $b,c,d,e$ must map onto $\{1,3,4\}$ using 4 elements, with one of $1,3,4$ repeated.

• Choose which of $1,3,4$ is repeated: $3$ ways.
• Choose the 2 elements (out of $b,c,d,e$) that map to that repeated value: $\binom{4}{2}=6$ ways.
• Assign the remaining two elements to the remaining two values: $2! =2$ ways.

So, $3\cdot 6\cdot 2 = 36.$

Case 2: Exactly one among $b,c,d,e$ maps to $2$

Then the remaining three elements must map bijectively to $1,3,4$.

• Choose which one of $b,c,d,e$ maps to $2$: $4$ ways.
• Assign the remaining three elements to $1,3,4$: $3! =6$ ways.

So, $4\cdot 6 = 24.$

Therefore, for fixed $f(a)=2$, total number of onto functions is $36+24=60.$

6. Since $f(a)$ can be any of $2,3,4$, by symmetry total number is $3\times 60 = 180.$

7. Hence the required number of onto functions is $\boxed{180}.$

8. Comparison with stored correct answer: Stored correct answer = $180$. This matches our derived answer.